Mensuration | ETE-889022705
Best Defence Institute for AFCAT , CDS , NDA , CAPF , ACC , TA Mob no :- 87-09-49-64-74 Page | 1
Mensuration
1. The perimeter of a right - angled triangle is 60
c.m. Its hypotenuse is 28 c.m. The area of the
triangle is :-
(1) 120 c.m.2 (2) 240 c.m.2
(3) 75 c.m.2 (4) 60 c.m.2
2. The cost of tiling a rectangular room is Rs.
12852 at the rate of Rs. 68 per square ft. The
length of the room is 8 ft. less than the side of a
square whose area is 841 square ft. What is the
breadth of the rectangular room?
(1) 8 ft. (2) 12 ft.
(3) 7 ft. (4) None of these
3. The area of a rectangle is equal to the area of a
circle with circumference equal to 52.8 meters.
What is the length of the rectangle if its breadth
is 5.5 meters?
(1) 40.32 m (2) 42.56 m
(3) 36.82 m (4) None of these
4. A rectangular carpet has an area of 176 sq.
meters and a perimeter of 52 meters. The length
of its diagonal is?
(1) 19 m (2) 22 m
(3) 16 m (4) None of these
5. The length of a rectangle is three times of its
breadth If the length of its diagonal is 7 c.m,
then the perimeter of the rectangle is?
(1) 48 cm. (2) 52 cm.
(3) 56 cm. (4) 64 cm.
6. A cuboid of dimension 27 cm. × 16 cm. × 15
cm. is melted and smaller cubes are of side 6
cm. is formed. Find how many such cubes can
be formed?
(1) 15 (2) 30
(3) 45 (4) 40
7. A circular wire of diameter 84 cm. is bent in the
form of a rectangle whose sides are in the ratio
13:9. The area of the rectangle is?
(1) 4096 cm2 (2) 4224 cm2
(3) 4212 cm2 (4) 4190 cm2
8. The area of a circle is 440 square cm. The area
of square inscribed in this circle will be?
(1) 320 cm.2 (2) 260 cm.2
(3) 280cm.2 (4) 240 cm2
9. A copper sphere of diameter 36 c.m is drown
into a wire of diameter 8 mm. Find the length of
the wire?
(1) 512 m (2) 486 m
(3) 408 m (4) 398 m
10. Find the difference between the volume of a
cylinder and volume of a cone having same
radius and same height. Radius is equal to length
of a rectangle having area 1120 cm2 and breadth
32 cm and height is half of the radius.
(1) 40001.01 cm3 (2) 44916.67 cm3
(3) 43333.33 cm3 (4) 51111.99 cm3
11. A solid cone of radius 21 cm and height 10/7 of
its radius is melted to form four solid cylinders
of radius 7 cm. Find the difference between the
height of the cone and height of each of the
solid cylinders.
(1) 9 cm (2) 6 cm
(3) 7.5 cm (4) 5.5 cm
12. A path of width 4m is running around a
rectangular field of length 24 m and breadth 18
m. Find the cost of flooring the path at the cost
of Rs.300/m2.
(1) Rs.60000 (2) Rs.90000
(3) Rs.120000 (4) Rs.100000
10
Mensuration | ETE-889022705
Best Defence Institute for AFCAT , CDS , NDA , CAPF , ACC , TA Mob no :- 87-09-49-64-74 Page | 2
13. The sum of areas of two rectangles (R1 and R2)
of same breadth is 720 cm2, and the ratio of
length of rectangle R1 and length of rectangle
R2 is 5:4 respectively. If the area of square
having length of side equal to the breadth of
rectangle is 256 cm2, then find the area of
rectangle R1.
(1) 320 cm2 (2) 400 cm2
(3) 240 cm2 (4) 480 cm2
14. If the length and breadth of a rectangular
seminar hall are each increased by 2m, then the
area of floor is increased by 42 sq.m. If the
length is increased by 2m and breadth is
decreased by 2m then the area is decreased by
10sq.m. The perimeter of the floor is
(1) 36m (2) 38m
(3) 34m (4) 32m
15. The difference between two parallel sides of
trapezium is 8cm. The perpendicular distance
between them is 38cm. if the area of the
trapezium is 950 cm^2, find the length of the
parallel sides.
(1) 27, 35 (2) 31, 40
(3) 29, 21 (4) 41, 49
16. The diameter of a rod is 2 cm and its length is
32 cm. This rod is drawn into a wire of length
128 cm and uniform thickness. What is the
thickness of the new wire?
(1) 2 cm (2) √2 cm
(3) 1 cm (4) 1.5 cm
17. Rajan is trying to make a cube using three old
cubes having edges 6 cm, 8 cm and 10 cm. What
will be the edge of new cube?
(1) 3 cm (2) 5 √2 cm
(3) 12 cm (4) 6 cm
18. The area of a rectangular parking space is 144
m2. If the length had been 6 meters more, the
area would have been 54 m2 more. The original
length of the parking space is
(1) 22 metres (2) 18 metres
(3) 16 metres (4) 24 metres
19. What is the cost of ploughing a land which
length and breadth in the ratio of 12:6 and its
perimeter is 840 feet. If the cost of needed to
plough a land is Rs.11 per 100 sq. feet.
(1) Rs.5648 (2) Rs.4312
(3) Rs. 6,660 (4) Rs.7,550
20. The barrel of a Ink pen is cylindrical in shape
which radius of base as 0.12 cm and is 7 cm
long. One such barrel in the pen can be used to
write 450 words. A barrel full of ink which has a
capacity of 17 cu. cm can be used to write how
many words approximately?
(1) 15423 (2) 21342
(3) 17645 (4) 24147
21. PQR is a right angle triangle with right angle at
Q. if the semi-circle on PQ with PQ as diameter
encloses on area of 243 sq.cm and the semicircle
on QR with QR as diameter encloses an area of
108 sq.cm then the area of the semicircle on PR
with PR as diameter will be.
(1) 305 sq.cm (2) 325 sq.cm
(3) 351 sq.cm (4) 362 sq.cm
22. The length and breadth of the hall are 60 feet
and 30 feet respectively. Square tiles of a 4 feet
length of different colors are to be laid on the
floor. Red tiles are laid in the 1st row on all
sides. If green tails are laid in the one – two time
of the remaining and yellow in the rest,
approximately how many yellow tiles will be
there?
(1) 52 (2) 63
(3) 72 (4) 35
23. A rectangular sheet, when folded into 2
corresponding parts had a perimeter 102cm for
each part folded along one set of sides and the
same is 114cm when folded along the other set
of sides. What is the area of the sheet?
(1) 1520 sq.cm (2) 1402 sq.cm
(3) 1310 sq.cm (4) 1260 sq.cm
24. Smallest angle of a triangle is equal to two-third
the smallest angle of a quadrilateral. The ratio
between the angles of the quadrilateral is
3:4:5:6. Largest angle of the triangle is twice its
smallest angle. What is the sum of second
largest angle of the triangle and largest angle of
the quadrilateral?
(1) 160º (2) 180º
(3) 190º (4) 170º
25. The volume ratio of Cylinder 1 to Cylinder 2 is
5:7 and its height is in the ratio of 2:3.Find the
Radius ratio of both the cylinders?
(1) 10:21 (2) 10 1/2 : 21 ½
(3) 15 1/2 : 14 ½ (4) 13:14
26. Amrita goes walking at morning. She walks
along the boundary of the walking park which is
in rectangular shape and she walks at a speed of
Mensuration | ETE-889022705
Best Defence Institute for AFCAT , CDS , NDA , CAPF , ACC , TA Mob no :- 87-09-49-64-74 Page | 3
8kmph completes her one round in 15 minutes.
Find the area of park, if the length and breadth is
in the ratio of 3:2?
(1) 240000 (2) 126500
(3) 365200 (4) 250000
27. The radius of the base of a cylindrical vessel is
21 cm and its height is 12 cm. The vessel is full
of milk. How many cylindrical vessels each
having diameter 14 cm and height 4 cm can be
filled with the milk in the big cylindrical vessel?
(1) 24 (2) 20
(3) 22 (4) 27
28. Length, breadth and height of a cuboidal box is
6 cm, 4 cm and 3 cm, respectively. A person
keeps 18 such box one above the other and then
paints it, and amount incurred was Rs. x. If all
boxes are painted separately then amount
incurred would be Rs. y. Find the difference
between ‘x’ and ‘y’ if the cost of painting the
box per cm2 is Rs. 2.
(1) Rs. 1520 (2) Rs. 1632
(3) Rs. 1266 (4) Rs. 1484
29. Length of a cuboid is 3 cm, and the ratio of its
breadth and height is 1:3. If the length of
diagonal is 13 cm2, then find the volume of the
cuboid.
(1) 184 cm3 (2) 164 cm3
(3) 254 cm3 (4) 144 cm3
30. A solid cone is cut vertically into two equal
parts. If radius and volume of the whole solid
cone before it is cut is 6 cm and 216√3 cm3,
respectively, then find the total surface area of
one of the cut portions.
[Use π = 3]
(1) 224.3 cm2 (2) 252.2 cm2
(3) 286.7 cm2 (4) 184.9 cm2
31. A cone is inscribed in a hollow cylinder. The
height and the base of cone and cylinder is the
same. The cost of painting the curved surface
area of the cylinder is Rs. 9072 at Rs. 14 per
cm2, and the curved surface area of cone is 405
cm2. Find the space remaining in the cylinder
after inscribing cone. [Take π = 3]
(1) 1902 cm3 (2) 1856 cm3
(3) 1920 cm3 (4) 1944 cm3
32. A piece of iron size, 14 inches in length, 11
inches in breadth and 3 inches in thick was
melted and remodified in the form of a rod of 14
inches diameter. Then find the length of the rod.
(1) 8 inches (2) 2 inches
(3) 7 inches (4) 3 inches
33. Radius of curvature at two points (A and B) of
spiral is given. What should be the
circumference of the spiral?
(1) 65. 84 cm (2) 62.80 cm
(3) 57.76 cm (4) 50.24 cm
34. Outside a rectangular garden a path of uniform
width of 3 meter is made. If the area of
rectangular garden is 640m2 and area of garden
including path is 916m2, then find the perimeter
of the garden.
(1) 30 m (2) 50 m
(3) 20 m (4) 80 m
35. The perimeter of a square is equal to the
perimeter of a rectangle of length 14 cm and
breadth 20 cm. Find the circumference of a
semicircle (approx.) whose diameter is equal to
the side of the square.
(1) 30.23cm (2) 43.71 cm
(3) 17.5cm (4) 50.47cm
36. Area of a rectangular plot is 75√3 cm2. If the
sum of its diagonals is four times the shorter
side, then the longer side would be?
(1) 12 cm (2) 15 cm
(3) 18 cm (4) 21 cm
37. A circle of maximum possible area is cut out
from a rectangular cardboard sheet of length 10
cm and breadth 6 cm. What is the remaining
area of the cardboard?
(1) 3(20 - 6π) cm2 (2) 3(20 - 3π) cm2
(3) 3(20 - 5π) cm2 (4) 3(20 - 2π) cm2
38. The ratio between sides of a square and an
equilateral triangle is 11:8. If the area of triangle
is 144√3 m2, find the perimeter of square
(1) 124 meter (2) 96 meter
(3) 132 meter (4) None of these
Mensuration | ETE-889022705
Best Defence Institute for AFCAT , CDS , NDA , CAPF , ACC , TA Mob no :- 87-09-49-64-74 Page | 4
39. A rectangle having length as 21 cm and breadth
as 10 cm is rotated about its longer edge as axis.
Find the volume of the solid generated.
(1) 72.07 cm3 (2) 167.05 cm3
(3) 214.06 cm3 (4) 102.51 cm3
40. A solid spherical ball of volume 38808 cm3 is
broken into two equal halves. Find the increase
in the total surface area of both hemispheres
with respect to the sphere (in cm2).
(1) 2772cm2 (2) 2277cm2
(3) 2727cm2 (4) 2722cm2
41. The sum of areas of two rectangles (R1 and R2)
of same length is 594 cm2, and the ratio of
breadth of rectangle R1 to breadth of rectangle
R2 is 5:4. If the area of square having length of
side equal to the length of rectangle is 484 cm2,
then find the area of rectangle R1.
(1) 240 cm2 (2) 280 cm2
(3) 330 cm2 (4) 360 cm2
42. Curved surface area of a cylindrical vessel is
1584 cm2. If the height of the vessel is 18 cm,
then find the cost of painting the total surface
area of the vessel at the rate of Rs. 2 per cm2.
(1) Rs. 5632 (2) Rs. 5864
(3) Rs. 5540 (4) Rs. 5780
43. The length and breadth of a rectangular park is
48 m and ‘x’ m, respectively. If the cost of
fencing the rectangular park at the rate of Rs.
14/m is Rs. 2352, then find the area of a square
field having side equal to ‘x’.
(1) 1296 m2 (2) 1024 m2
(3) 1600 m2 (4) 1225 m2
44. If the volume of the cylindrical tank of height 14
cm and radius ‘r’ cm is 6336 cm3, then find the
value of ‘r’.
(1) 9 cm (2) 10 cm
(3) 12 cm (4) 15 cm
45. Length and breadth of a rectangular park is 44 m
and 38 m, respectively. A path is made around
the park from outside of width 3.5 m and paved
with marbles. Find the cost of paving the path, if
it costs Rs. 2 per cm2.
(1) Rs. 1142 (2) Rs. 1870
(3) Rs. 1246 (4) Rs. 1484
46. A took 20 seconds to cross a rectangular field
diagonally walking at the rate of 68 m/min and
B took the same time to cross the same field
along its side walking at the rate of 50 m/min.
The area of the rectangular field is?
(1) 234 m² (2) 260 m²
(3) 216 cm² (4) None of these
47. If a is the area, b is the circumference and c is
the diameter of circle then the value of a/bc
(1) 4 : 1 (2) 1 : 4
(3) 1 : 2 (4) 2 : 1
48. The radius and height of a right circular cone are
in the ratio 5:12. If its volume is 800 π cm3,
what is its slant height?
(1) 9 cm (2) 13 cm
(3) 26 cm (4) 17
49. Area of the square is equal to the area of the
rectangle. The ratio of the length and breadth of
the rectangle is 4:1 and the perimeter of the
square is 64 cm. Find the perimeter of the
rectangle
(1) 64 cm (2) 48 cm
(3) 72 cm (4) 80 cm
50. A road of width 7m is running around a circular
field from outside. The radius of the circular
field is 28m. Find the cost of fencing the road at
the rate of Rs.16/m2.
(1) Rs.18564 (2) Rs.22456
(3) Rs.12986 (4) Rs.22176
Mensuration | ETE-889022705
Best Defence Institute for AFCAT , CDS , NDA , CAPF , ACC , TA Mob no :- 87-09-49-64-74 Page | 5
Mensuration (Solution)
Ans.1(4) Let base = b cm. Height = h c.m.
b + h + 28 = 60
b + h = 32
(b+h)2 = (32)2..........(I)
Also, b2 + h2 = 282
2bh = 60×4
bh = 120
bh = 60
Area = 60 c.m.2
Ans.2(4) The area of rectangular room = Sq.ft
The side of the square room = = 29 ft
length = 29-8 = 21 ft
breadth = = 9 ft .
Ans.3(1) Circumference of the circle = 52.8 m
2r = 52.8
2× ×r = 52.8
r = 8.4 m
Area of the circle = r2
×8.4×8.4 = 221.76 m2
Area of rectangle
Length of the rectangle = = 40.32m
Ans.4(4) 2(l+b)=52
l+b = 26 , and lb = 176 m2
Diagonal =
=
= = = 18 m
Ans.5(3) Let breadth = x cm then, length = 3x cm
x2+(3x)2 =
10x2= 490
x = 7 cm.
then, length = 21 cm.
bredth = 7 cm.
Perimeter = 2 (l+b)
= 2 (7+21)
= 56 c.m
Ans.6(2) No. of cubes =
Ans.7(3) Length of wire = 2x =264 cm.
Perimeter of rectangle = 2(13x+9x)=264
cm.
= 44x=264
x=6
Area of Rectangle = (13x6)x(9x6)
= 4212 cm2
Ans.8(3) R2 = 440 = R2=440× = 140
So, R = ×(Diagonal)
Diagonal = 2 R
Area of the square = ×(Diagonal)2
6= ×4R2 = 2R2
= 2×140
= 280 cm2
Ans.9(2) Volume of sphere = × × ×
= 7776 cm3
Volume of wire = ( x0. 4×0.4×h) cm2
7776 = × × × h
h = 48600 cm
h = m
h = 486 m
Ans.10(2) Area of rectangle = length x breadth
=> 1120 = length x 32
=> Length = 1120/32
=> Length = 35 cm
Radius = 35 cm
Height = 35/2 cm
Volume of cylinder – Volume of cone
= πr2h – 1/3 πr2h
= πr2h x (3 – 1)/3
= 2/3 x πr2h
= 2/3 x 22/7 x 35 x 35 x 35/2
= 44916.67 cm3
Ans.11(3) We know that volume of cone
= 1/3πr2h
And volume of cylinder = πr2h
1/3 x 22/7 x 21 x 21 x 10/7 x 21 = 4 x
22/7 x 7 x 7 x h
= h = 1/3 x 22/7 x 21 x 21 x 30 x ¼ x 7/22
x 1/7 x 1/7
= h = 22.5 cm
Required difference = 10/7 x 21 – 22.5
= 30 – 22.5
= 7.5 cm
Ans.12(3)
Area of the path = (32 x 26) – (24 x 18)
2 2 2 2 2( ) 32 28b h b h
1
2
12852
68
841
189
21
227
227
221.76
5.5
2 2 2l b (l b) 2lb
226 2x176
676 352 324
2
7 10
27x16x1530
6x6x6
22 84
7 2
π
7
22
1
2
1
2
4
3
36
2
36
2
36
2
4
10
4
10
48600
100
Mensuration | ETE-889022705
Best Defence Institute for AFCAT , CDS , NDA , CAPF , ACC , TA Mob no :- 87-09-49-64-74 Page | 6
= 832 – 432
= 400 m2
Cost of flooring the path = 400 x 300
= Rs.120000
Ans.13(2) Let, length of rectangle R1 and rectangle
R2 be 5x cm and 4x cm respectively.
Breadth of rectangle = side of square
= √256 = 16 cm
According to question,
Sum of areas of two rectangles (R1 and
R2)
= 720
5x × 16 + 4x × 16 = 720
16(5x + 4x) = 720
16x = 720/9
x = 80/16, x = 5
So, the length of rectangle R1 = 5 × 5
= 25 cm
Therefore, area of rectangle R1 = 25 × 16
= 400 cm2
Ans.14(2) Let length = x metres
Breadth = y metres
Then (x+2) (y+2) – xy = 42
Xy+2x+2y+4 – xy = 42
2x +2y = 42 – 4
2x +2y = 38
X+y = 19 ------------->1
And xy – [(x+2)(y-2)] = 10
Xy – [ xy – 2x + 2y +4)] = 10
Xy – xy +2x – 2y + 4 = 10
2x – 2y =6
X – y = 3 ---------------->2
Adding (1) & (2)
2x = 22
X=11
Put x value in (1)
Y = 19 -11
=8
So length = 11m and breadth = 8m
Perimeter = 2(l+b)
=2(11+8)
=38m
Ans.15(3) Let the 2 parallel sides of the trapezium be
x cm and y cm
Then x – y = 8 ========>1
½ (x+y)× 38 = 950
19 (x+y) = 950
X+y = 950/19
X+ y = 50 ==========>2
Solving 1 and 2
2x = 58
x= 29, y = 21
So the parallel sides are 29cm and 21 cm.
Ans.16(3) πR2H = πr2h
12 × 32 = r2 × 128
r2 = ¼
r = ½
Diameter = 1 cm
Ans.17(3) It has been given that the new cube is
made of old cubes. Volume of new
cubes, 6 cm, 8 cm And 10 cm The sum of
the volume of the cube with the sides will
be equal to.
Volume of a cube ( =side)3
Thus, the volume of the new cube
= (6)3 + (8)3 + (10)3
⇒ Volume = 216 + 512 + 1000
= 1728cm3
∴ Side of 1728 cubic cm volume
=∛(volume) = √1728 =12 cm. Ans.18(3) l × b = 144
(l + 6) × b = 144+54 = 198
So, (l + 6)/l = 198/144
1 + 6/l = 11/8
6/l = 11/8 – 1 = 3/8
l = 16 m
Ans.19(2) Let length be l and breadth be b.
Then the perimeter of the floor = 2 (l+b)
2(l+b) = 840
l+b =420
l = 12/18 ×420 =280 feet
b= 6/18 × 420 = 140 feet
Area = 140×280 = 39200 sq.feet
Cost of laying = 392×11 = Rs.4312
Ans.20(4) Volume of the barrel of Ink pen = πr2h =
22/7 × 0.12×0.12 × 7 = 0.3168cu cm
A barrel which has capacity 0.3168 cu. cm
can write 450 words
So which has capacity 17 cu cm can write
= 450/0.3168×17= 24147 words
Ans.21(3) Required area = π/2 (PR/2)2
=π/2 (PR)2/4
=π/2 [(PQ2+QR2)/4]
=π/2 [PQ2/4 + QR2/4]
=π/2 (PQ2/4) + π/2 (QR2/4)
=243 +108
=351 sq.cm
Ans.22(4) Area left after laying red tiles = [(60-
8)×(30-8)]
= 52×22
= 1144 sq.ft
Area under green tiles = 1144 / 2 = 572
sq.ft
Area under yellow tiles = 1144-572=572
sq.ft
Number of yellow tiles = 572/ 16
= 35
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Best Defence Institute for AFCAT , CDS , NDA , CAPF , ACC , TA Mob no :- 87-09-49-64-74 Page | 7
Ans.23(4) When folded along breadth, we have 2
(l/2 + b) = 102
L + 2b = 102 ---------------->1
When folded along length, we have 2
(l+b/2) = 114
2l+b = 114 ------------------>2
Multiply (2) by 2, we get
4l + 2b =228 --------------->3
Subtract 1 from 3
3l = 126 => l=42
Put l value in (1) we get b = 30
Area of the sheet = (42×30) cm2
=1260 sq.cm
Ans.24(2) Let be ratio = x
Sum of angles of the quadrilateral
⇒ 3x + 4x + 5x + 6x = 360o
⇒ 18 x = 360o
⇒ x = 20o
Angles of quadrilateral = 60o, 80o, 100o,
120o
Smallest angle of triangle = 40o
Largest angle of triangle = 80o
Sum = 60o + 120o
=180o
Ans.25(3) πR2h1 / πr2h2 = 5/7
R2 / r2 3 =5/7
R/r =15 1/2 : 14 1/2
Ans.26(1) Circumference = (8000/60)×15
= 2000m
2(l+b) = 2000
l+b =1000
3x+2x =1000
5x= 1000
x=200
Area = 3x×2x= 600×400 =240000m2.
Ans.27(4) Radius of each of the small cylindrical
vessels
= 14/2 = 7 cm
Let, required number of cylindrical
vessels = n
We know that,
Volume of cylinder = πr2h
According to the question
22/7 x 21 x 21 x 12 = n x 22/7 x 7 x 7 x 4
=> n = 22/7 x 21 x 21 x 12 x 7/22 x 1/7 x
1/7 x ¼
=> n = 27
Ans.28(2) Total surface area of each box
= 2 × [(6 × 4) + (4 × 3) + (3 × 6)]
= 2 × (24 + 12 + 18) = 108 cm2
Total cost incurred for painting 18 boxes
separately = 2 × 18 × 108 = Rs. 3888
So, y = 3888
Total surface area ifall boxes are kept one
above the other = 2 × [(6 × 4) + (4 × 54) +
(54 × 6)] = 2 × (24 + 216 + 324)
= 1128 cm2
Total cost incurred for painting the boxes
that are kept one above the other
= 2 × 1128 = Rs. 2256
So, x = 2256
Required difference = 3888 – 2256
= Rs. 1632
Ans.29(4) Let breadth and height of the cuboid be x
cm and 3x cm, respectively.
According to question,
x2 + (3x)2 + 32 = 132
x2 + 9x2 + 9 = 169
10x2 = 160
x2 = 16
x = 4 cm
Breadth of cuboid = 4 cm
Height of cuboid = 12 cm
Volume of cuboid = 3 × 4 × 12 = 144 cm3
Ans.30(1) Let, height of cone = ‘h’ cm
Given, volume of the solid cone
= 216√3 cm3
Radius of cone = 6 cm
So, 1/3 × π r2 h = 216√3
1/3 × 3 × 6 × 6 × h = 216√3
h = 6√3 cm
Diameter of base circular part of solid
cone
= 12 cm
Slant height of solid cone = √{(6√3)2 +
62}
= √144 = 12 cm
So, total curved surface area of the cut
portion
= (πrl/2) + (πr2/2) + (Area of the
equilateral triangle with side 12 cm)
= (3 × 6 × 12)/2 + (3 × 62)/ 2 + (√3/4 ×
122)
= 108 + 54 + 36√3
= (162 + 36√3) cm2
= 224.3 cm2
Ans.31(4) Let, radius and height of cone and
cylinder be r cm and h cm, respectively.
And, slant height of cone = l cm
Curved surface area of cylinder
= 9072/14 = 648 cm2
So, 2 × 3 × r × h = 648
rh = 108
r = 108/h --------- (i)
And, 3 × r × l = 405
r = 135/l --------- (ii)
From (i) and (ii)
Mensuration | ETE-889022705
Best Defence Institute for AFCAT , CDS , NDA , CAPF , ACC , TA Mob no :- 87-09-49-64-74 Page | 8
108/h = 135/l
l/h = 135/108
l/h = 5/4
Let, slant height and height of cone be 5x
cm and 4x cm, respectively.
So, 25x2 = 16x2 + r2
9x2 = r2
r = 3x
So, 3x = 108/4x
12x2 = 108
x2= 9
x = 3
Height of cone = 4x = 12 cm
Required space = 3 × 9 × 9 × 12 – 1/3 × 3
× 9 × 9 × 12 = 1944 cm3
Ans.32(4) Given that diameter= 14/2= 7 inches
According to the question,
14×11×3= 22/7×7×7×l
462=154×l
l= 3 inches
Ans.33(3)
Minimum and maximum value of the
circumference of the spiral will be the
circumferences of the circles given in the
diagram
Circumference of smaller circle
= 2⨉3.14⨉8 = 50.24 cm
Circumference of bigger circle
= 2⨉3.14⨉10 = 62.80 cm
50.24 < Circumference of the spiral <
62.80
Thus option C is the valid answer
Ans.34(4) Let length and Width of the inner
boundary be x and y respectively
Given that area of the garden=640m2
(i.e) xy=640m2……………… (1)
Now length of outer boundary path=x+6
And width of outer boundary path=y+6
Hence Area of garden including path
=(x+6)(y+6)=916
(i.e) xy+6x+6y+36=916………………….
(2)
From (1) and (2), we have
640+6x+6y+36=916
x+y=40
Thus perimeter of the
garden=2(x+y)=80m
Ans.35(2)
Ans.36(2) 𝑙 × 𝑏 = 75√3 - - - - (1)
2√𝑙2 + 𝑏2 = 4𝑏
𝑙2 + 𝑏2= 4𝑏2
𝑙2 = 3𝑏2
𝑏 = 𝑙/√3 Putting this value in equation 1.
𝑙 × 𝑙/√3 = 75√3
𝑙2 =225
𝑙 = 15 Ans.37(2) A circle cut out from the rectangular
cardboard can attain a maximum diameter
of 6cm.
So, maximum possible area of the circle
= π × 32 = 9π cm2
Area of the rectangular cardboard
= 10×6 = 60 cm2
Therefore, remaining area of cardboard
= 60 - 9π cm2 = 3(20 - 3π) cm2
Hence, option 2.
Ans.38(3) Let the side of a triangle = x meter
√3
4 x2 = 144√3
x2 = 144×4
x = 24 meter
perimeter of square = 33×4 = 132 meter
Ans.39(2) As the rectangle is rotated about its longer
edge, 21 cm becomes the height of the
cylinder while 10 cm becomes the
circumference of the base of the cylinder.
2 × π × r = 10
So, r = 35/22 cm
Thus, volume of cylinder = π × r2 × h
= (22/7) × (35/22) × (35/22) × 21
= 3675/22 cm3 = 167.05 cm3
Ans.40(1) Let the radius of the sphere be ‘r’ cm.
Given, Volume of sphere = 38808 cm3
(4/3) × π × r3 = 38808
r3 = 9261
r = 21 cm
Now, total surface area of sphere before
breaking into two halves = 4πr2
Total surface area of two hemispheres
= 2 × 3πr2 = 6πr2
Thus, increase in total surface area
= 6πr2– 4πr2 = 2πr2
= 2772 cm2
Ans.41(3) Let, breadth of rectangle R1 and rectangle
R2 be 5x cm and 4x cm respectively.
Length of rectangle = side of square
= √484 = 22 cm
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Best Defence Institute for AFCAT , CDS , NDA , CAPF , ACC , TA Mob no :- 87-09-49-64-74 Page | 9
According to question,
Sum of areas of two rectangles (R1 and
R2)
= 594
22 × 5x + 22 × 4x = 594
22(5x + 4x) = 594
9x = 594/22
9x = 27, x = 27/9 = 3
So, the breadth of rectangle R1= 5 × 3
= 15 cm
Therefore, area of rectangle R1
= 22 × 15 = 330 cm2
Ans.42(1) Let, radius of vessel be ‘r’ cm.
So, 2 × 22/7 × r × 18 = 1584
r = 14 cm
Required cost = 2 × 22/7 × 14 × (14 + 18)
×2
= Rs. 5632
Ans.43(1) So, 2 × (48 + x) × 14 = 2352
48 + x = 84
x = 84 – 48 = 36
So, area of square field = 36 × 36
= 1296 m2
Ans.44(3) Volume of cylinder = π r2 h
6336 = 22/7 × r2 × 14
r2 = 144
r = 12 cm
So, the value of r = 12 cm
Ans.45(3) Area of path paved with marble
= (44 + 3.5 × 2) × (38 + 3.5 × 2) – 44 × 38
= 2295 – 1672 = 623 cm2
Required cost = 623 × 2 = Rs. 1246
Ans.46(4)
Ans.47(2) We know that,
Area of a circle = πd2 /4
Where, d = Diameter of the circle
⇒ a = πc2 /4
Also, Circumference of the circle = πd
⇒ b = πc
a/bc = (πc2 /4)/πc2 = 1 : 4
Ans.48(3) As per the given data
r/h = 5/12
⇒ h = 12r/5
Also given that volume of the cone is 800
π cm 3
⇒ 1/3 π r2 h = 800 π
⇒ 1/3 × r2x12r/5 = 800
⇒ 12r3 /15 = 800
⇒r3= 1000
⇒ r = 10 cm
∴height of the cone = 12r/5 = 12 (10)/5 =
24 cm
∴slant height of the cone = √(242 + 102) =
26 cm
Ans.49(4) Side of a square = 4a =64
=> a= 64/4
= 16 cm
Area of a rectangle = Area of the square
=16×16 = 256
(i.e) Area of a rectangle =256
(i.e) 4x×x = 256
=> x2 = 256/4
= 64
=> x= 8 cm
Perimeter of the rectangle = 2 (l+b) = 2
(4x+x)
= 2 (4×8+8)
= 2×40
= 80 cm
Ans.50(4) Area of the circular field = π(R2 – r2)
= [22/7 x (28 + 7)2 ]– [22/7 x (28)2]
= 22/7 x (35)2 – 22/7 x (28)2
= 3850 – 2464
= 1386 m2
Cost of fencing = 16 x 1386 = Rs.22176
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